Question

A galvanic cell consists of an electrode of Pb in a 0.03 M solution of PbSO4 and an electrode of Sn in a solution of 0.7 M SnSO4 at 25°C. What is the emf of the cell?

Answer 1

Oxidation : Sn ----------> Sn+2 +   2e-     Eo = - 0.14 V

Reduction : Pb+2 + 2e-    --------------> Pb    Eo = - 0.13 V

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overall : Sn (s) + Pb+2 (aq)   -----------> Sn+2 (aq) + Pb (s)

Eocell = - 0.13 - (- 0.14)

          = 0.01 V

Ecell = Eo - 0.05916 / n log [Sn+2 / Pb+2]

         = 0.01 - 0.05916 / 2 log [0.7 / 0.03]

Ecell = - 0.0305 V